Spatial Search and Commuting with Asymmetric Changes of the Wage Distribution

Abstract

This paper contributes to job-search literature by analysing commuter behaviour in the presence of asymmetric changes in the wage distribution. Job search theory predicts that reservation wages increase with the mean and mean-preserving spread of the wage distribution. However, changing dispersion while holding the mean constant implies symmetric stretching or compression of the wage distribution in both tails, which is not likely to be the case when confronted with the real data. The presented model predicts that the commuter stream and the reservation wage increase with the median-preserving spread in the right tail and decrease with the median-preserving spread in the left tail in the destination. The empirical part, based on German commuter data, confirms the theory’s predictions.

Appendix

1.1 Proof of Proposition 1

To derive the effect of the change of the median on the reservation wage one needs to introduce the notion of the translation of the distribution. Changing the median while holding the shape of the distribution constant is simply a parallel shift of the distribution. If we increase the median of the distribution $F(x)$ by the value $\mu$, the resulting distribution would be a translation of the original c.d.f. $F(x)$. The distribution $G(x)$ is a translation of $F(x)$ if $G(x + \mu ) = F(x)$ and, hence, $G(x) = F(x - \mu )$.

The reservation wage in region A is given as:

$ w_A^R = b - {c_A}\left( {{\theta_A}} \right) - {c_B}\left( {{\theta_B}} \right) + \frac{{{\lambda_A}\left( {{\theta_A}} \right)}}{r}\int\limits_{w_A^R}^\infty {\left( {w - w_A^R} \right){\text{d}}{F_A}(w)} + \frac{{{\lambda_B}\left( {{\theta_B}} \right)}}{r}\int\limits_{w_B^R}^\infty {\left( {w - w_B^R} \right){\text{d}}{F_B}(w)} $

((10))

Increase of the median in region A by $\mu$ (given that $w_B^R = w_A^R + \delta$) would result in a new reservation wage:

$ w_A^R(\mu ) = b - {c_A}\left( {{\theta_A}} \right) - {c_B}\left( {{\theta_B}} \right) + \frac{{{\lambda_A}\left( {{\theta_A}} \right)}}{r}\int\limits_{w_A^R(\mu )}^\infty {\left( {w - w_A^R(\mu )} \right){\text{d}}{F_A}(w - \mu )} + \frac{{{\lambda_B}\left( {{\theta_B}} \right)}}{r}\int\limits_{w_A^R(\mu ) + \delta }^\infty {\left( {w - w_A^R(\mu ) - \delta } \right){\text{d}}{F_B}(w)} $

((11))

Subtracting (10) from (11) we obtain:

$ \begin{array}{c} \left( w_A^R(\mu ) - w_A^R \right) = \frac{{\lambda_A\left( \theta_A \right)}}{r} \left[ \int\limits_{w_A^R(\mu )}^\infty {\big(w - w_A^R(\mu )\big){\text{d}}{F_A}} (w - \mu ) - \int\limits_{w_A^R}^\infty {(w - w_A^R){\text{d}}{F_A}} (w)\ \right]\cr \quad + \\ \frac{{{\lambda_B}\left( {{\theta_B}} \right)}}{r}\left[ {\int\limits_{w_A^R(\mu ) + \delta }^\infty {\big(w - w_A^R(\mu ) - \delta \big){\text{d}}{F_B}} (w) - \int\limits_{w_A^R + \delta }^\infty {(w - w_A^R - \delta ){\text{d}}{F_B}} (w)} \ \ \right]\end{array}$

((12))

By integration by parts one obtains:

$ \int\limits_{w_A^R(\mu )}^\infty {\left(w - w_A^R(\mu )\right){\text{d}}{F_A}} (w - \mu ) = E(w) + \mu - w_A^R(\mu ) + \int\limits_0^{w_A^R(\mu )} {{F_A}} (w - \mu ){\text{d}}w $

((13))

and

$ \int\limits_{w_A^R}^\infty {\left(w - w_A^R\right){\text{d}}{F_A}} (w) = E(w) - w_A^R(\mu ) + \int\limits_0^{w_A^R} {{F_A}} (w - \mu ){\text{d}}w $

((14))

Dividing (11) by $\mu$ and taking the limit at $\mu = 0$ with the help of the results obtained in (13) and (14) one gets:

$ \frac{{\partial w_A^R}}{\partial \mu } = \begin{array}{c} \frac{{{\lambda_A}\left( {{\theta_A}} \right)}}{r}\left[ {1 - \frac{{\partial w_A^R(\mu )}}{{\partial \mu }}\left( {1 - {F_A}(w_A^R)} \right) - {F_A}(w_A^R)} \right] \cr \quad + \frac{{{\lambda_B}\left( {{\theta_B}} \right)}}{r}\left[ { - \frac{{\partial w_A^R(\mu )}}{{\partial \mu }}\left( {1 - {F_B}(w_A^R + \delta )} \right)} \right] \end{array} $

((15))

Therefore:

$ \frac{{\partial w_A^R}}{{\partial \mu }} = \frac{{{\lambda_A}\left( {{\theta_A}} \right)\left( {1 - {F_A}(w_A^R)} \right)}}{{r + {\lambda_A}\left( {{\theta_A}} \right)\left( {1 - {F_A}(w_A^R)} \right) + {\lambda_B}\left( {{\theta_B}} \right)\left( {1 - {F_B}(w_A^R + \delta )} \right)}} $

((16))

Obviously, $0 \leq \frac{{\partial w_A^R}}{{\partial \mu }} \leq 1$. Knowing that $w_A^R + \delta = w_B^R$, one obtains $\frac{{\partial w_A^R}}{{\partial \mu }} = \frac{{\partial w_B^R}}{{\partial \mu }}$. In the same fashion, increasing the median in B by $\mu$:

$ \frac{{\partial w_B^R}}{{\partial \mu }} = \frac{{{\lambda_B}\left( {{\theta_B}} \right)\left( {1 - {F_B}(w_B^R)} \right)}}{{r + {\lambda_B}\left( {{\theta_B}} \right)\left( {1 - {F_B}(w_B^R)} \right) + {\lambda_A}\left( {{\theta_A}} \right)\left( {1 - {F_A}(w_A^R)} \right)}} $

((17))

Again, $0 \leq \frac{{\partial w_B^R}}{{\partial \mu }} \leq 1$.

For the effect of the spreads, assume that the reservation wage is below the median (reservation wages above the median would be unaffected by the spread below the median). Denote $\Lambda = \int\limits_{w^R}^\infty {(w - w^R){\text{d}}F(w)}$. One could rewrite:

Fig. 2

Distribution of spreads in the right tail across regions. The ratios of the spreads in region N to the spreads in region N-1 are depicted on the axes. So if the change (or variation) of the spread in the left tail from region to region were equal to the change of the spread in the right tail (which symmetric stretching or compression would imply), the 45-degree line would have been observed. The graph shows that in real data the variation in spreads is largely asymmetric

$ \Lambda = \int\limits_{w^R}^{\bar w} {(w - w^R){\text{d}}F(w)} + \int\limits_{\bar w}^\infty {(w - w^R){\text{d}}F(w)} \hfill \\ = \frac{{\bar w}}{2} - \int\limits_{w^R}^{\bar w} {F(w){\text{d}}w} + \int\limits_{\bar w}^\infty {w{\text{d}}F(w)} - {w^R} \hfill \\ $

((18))

Note that $\frac{\partial }{{\partial \bar w}}\int\limits_{\bar w}^\infty {w{\text{d}}F(w)} > 0$. This result is intuitively clear – truncated mean increases if you move the truncation point to the right. Moreover, $\frac{\partial }{{\partial {\sigma_R}}}\int\limits_{\bar w}^\infty {w{\text{d}}F(w)} > 0$ - truncated mean increases if you increase the variance to the right of the truncation point. Hence, $\frac{{\partial \Lambda }}{{\partial {\sigma_R}}} > 0$. The effect of the spread in the left tail is: $\frac{{\partial \Lambda }}{{\partial {\sigma_L}}} = - \int\limits_{{w^R}}^{\bar w} {\frac{\partial }{{\partial {\sigma_L}}}F(w){\text{d}}w} < 0$, because $\frac{{\partial F(w)}}{{\partial {\sigma_L}}} > 0$. The logic here is straightforward – increasing the spread in the left tail moves some of the probability mass away to the left of the reservation wage (fewer jobs become attractive). As a result, the reservation wage declines to compensate for the loss of the probability mass. Hence, $\frac{{\partial w_A^R}}{{\partial {\sigma_{AR}}}} > 0$ and $\frac{{\partial w_A^R}}{{\partial {\sigma_{AL}}}} < 0$. In the same fashion one obtains $\frac{{\partial w_B^R}}{{\partial {\sigma_{BR}}}} > 0$ and $\frac{{\partial w_B^R}}{{\partial {\sigma_{BL}}}} < 0$. Given the relationship $w_A^R + \delta = w_B^R$, one also obtains: $\frac{{\partial w_A^R}}{{\partial {\sigma_{BR}}}} > 0$ and $\frac{{\partial w_A^R}}{{\partial {\sigma_{BL}}}} < 0$, and $\frac{{\partial w_B^R}}{{\partial {\sigma_{AR}}}} > 0$ and $\frac{{\partial w_B^R}}{{\partial {\sigma_{AL}}}} < 0$.

1.2 Proof of Proposition 2

Differentiate ${c'_A}\left( {{\theta_A}} \right) = \frac{{{{\lambda '}_A}\left( {{\theta_A}} \right)}}{r}\int\limits_{w_A^R}^\infty {\left( {w - w_A^R} \right){\text{d}}{F_A}(w)}$ with respect of ${\Lambda_A}$:

$ {c''_A}({\theta_A})\frac{{\partial {\theta_A}}}{{\partial {\Lambda_A}}} = \frac{{{{\lambda ''}_A}({\theta_A})}}{r}\frac{{\partial {\theta_A}}}{{\partial {\Lambda_A}}}{\Lambda_A} + \frac{{{{\lambda '}_A}({\theta_A})}}{r}, $

((19))

and

$ \frac{{\partial {\theta_A}}}{{\partial {\Lambda_A}}} = \frac{{{\lambda_A}({\theta_A})}}{{r{{c''}_A}({\theta_A}) - {{\lambda ''}_A}({\theta_A}){\Lambda_A}}} > 0. $

((20))

From $\frac{{\partial {\theta_A}}}{{\partial {{\bar w}_A}}} = \frac{{\partial {\theta_A}}}{{\partial {\Lambda_A}}} \cdot \frac{{\partial {\Lambda_A}}}{{\partial \bar w}}$ it immediately follows that $\frac{{\partial {\theta_A}}}{{\partial {{\bar w}_A}}} > 0,{\text{ }}\frac{{\partial {\theta_A}}}{{\partial {\sigma_{AR}}}} > 0,{\text{ and }}\frac{{\partial {\theta_A}}}{{\partial {\sigma_{AL}}}} < 0$. Differentiate ${c'_B}\left( {{\theta_B}} \right) = \frac{{{{\lambda '}_B}\left( {{\theta_B}} \right)}}{r}\int\limits_{w_B^R}^\infty {\left( {w - w_B^R} \right){\text{d}}{F_B}(w)}$ with respect to ${\bar w_A}$:

$ {c''_B}({\theta_B})\frac{{\partial {\theta_B}}}{{\partial {{\bar w}_A}}} = \frac{{{{\lambda ''}_B}({\theta_B})}}{r}\frac{{\partial {\theta_B}}}{{\partial {{\bar w}_A}}}{\Lambda_B} - \frac{{\partial w_B^R}}{{\partial {{\bar w}_A}}}\left( {1 - {F_B}(w_B^R)} \right)\frac{{{{\lambda '}_B}({\theta_B})}}{r}, $

((21))

and hence, $\frac{{\partial {\theta_B}}}{{\partial {{\bar w}_A}}} = \frac{{\frac{{\partial w_B^R}}{{\partial {{\bar w}_A}}}\left( {1 - {F_B}(w_B^R)} \right){{\lambda '}_B}({\theta_B})}}{{r{{c''}_B}({\theta_B}) - {{\lambda ''}_B}({\theta_B}){\Lambda_B}}} < 0$. In the same fashion: $\frac{{\partial {\theta_B}}}{{\partial {\sigma_{AR}}}} = \frac{{\frac{{\partial w_B^R}}{{\partial {\sigma_{AR}}}}\left( {1 - {F_B}(w_B^R)} \right){{\lambda '}_B}({\theta_B})}}{{r{{c''}_B}({\theta_B}) - {{\lambda ''}_B}({\theta_B}){\Lambda_B}}} < 0$ and $\frac{{\partial {\theta_B}}}{{\partial {\sigma_{AL}}}} = - \frac{{\frac{{\partial w_B^R}}{{\partial {\sigma_{AL}}}}\left( {1 - {F_B}(w_B^R)} \right){{\lambda '}_B}({\theta_B})}}{{r{{c''}_B}({\theta_B}) - {{\lambda ''}_B}({\theta_B}){\Lambda_B}}} > 0$.

If we want to see how search intensities react to changes in wages in both regions simultaneously, we simply let the wage distributions in both regions be identical. Then an increase in wages in region A would mean the same increase in region B. Then, ${c'_A}\left( {{\theta_A}} \right) = \frac{{{{\lambda '}_A}\left( {{\theta_A}} \right)}}{r}\int\limits_{w_A^R}^\infty {\left( {w - w_A^R} \right){\text{d}}{F}(w)}$ and ${c'_B}\left( {{\theta_B}} \right) = \frac{{{{\lambda '}_B}\left( {{\theta_B}} \right)}}{r}\int\limits_{w_B^R}^\infty {\left( {w - w_B^R} \right){\text{d}}{F}(w)}$. It is then easy to show that $\frac{{\partial {\theta_A}}}{{\partial \bar w}} > 0,{\text{ }}\frac{{\partial {\theta_A}}}{{\partial {\sigma_R}}} > 0{\text{, }}\frac{{\partial {\theta_A}}}{{\partial {\sigma_L}}} < 0$ and $\frac{{\partial {\theta_B}}}{{\partial \bar w}} > 0,{\text{ }}\frac{{\partial {\theta_B}}}{{\partial {\sigma_R}}} > 0{\text{, }}\frac{{\partial {\theta_B}}}{{\partial {\sigma_L}}} < 0$.

1.3 Proof of Proposition 3

The participation rate is given as $G(\bar b)$. Hence, the participation rate is increasing in $\bar b$. Since:

$ w_A^R(\bar b) = \bar b - {c_A}\left( {{\theta_A}} \right) - {c_B}\left( {{\theta_B}} \right) + \frac{{{\lambda_A}\left( {{\theta_A}} \right)}}{r}\int\limits_{w_A^R}^\infty {\left( {w - w_A^R} \right){\text{d}}{F_A}(w)} + \frac{{{\lambda_B}\left( {{\theta_B}} \right)}}{r}\int\limits_{w_B^R}^\infty {\left( {w - w_B^R} \right){\text{d}}{F_B}(w)}, $

((22))

and $w_A^R(\bar b) \equiv \bar b$, it is easy to show that $w_A^R(\bar b)$ and hence $\bar b$ increase with the median wage (both in the origin and destination) and the median-preserving spread in the right tail (both in the origin and destination), and decreases with the median-preserving spread in the left tail of the wage distribution (both in the origin and destination) (see also proof of Proposition 1).

Increase in the median wage in the destination would increase $\bar b$. Moreover, increase in the median in region B would make agents reallocate their intensity to region B (see Proposition 2) and therefore ${\lambda_B}({\theta_B})$ also increases. The reservation wage in B increases with the median wage in B but the elasticity is less than unity, hence, $(1 - {F_B}(w_B^R))$ also increases with the median wage in B. As a consequence, the commuter flow from A to B unambiguously increases with the median wage in the destination. If the median wage in the origin increases, the participation rate also goes up ($\bar b$ increases). However, agents would reallocate their search intensity to region A and hence, ${\lambda_B}({\theta_B})$ declines. This implies that when wages increase in the origin, less commuting is possible because agents start searching harder in the origin and less harder in the destination, but on the other hand, more commuting is possible because overall number of searchers in the origin increases. Hence, the overall effect is ambiguous.

To derive the effects of the spreads, consider first the case with exogenous search intensity. Denote $\phi = \lambda \int\limits_0^{\bar b} {\left( {1 - F\left( {{w^R}(b)} \right)} \right)} dG(b) = \lambda G\left( {\bar b} \right) - \lambda \int\limits_0^{\bar b} {F\left( {{w^R}(b)} \right)d} G(b)$. Moreover, integrating by parts, $\int\limits_0^{\bar b} {F\left( {{w^R}(b)} \right)d} G(b) = G\left( {\bar b} \right)F\left( {{w^R}(\bar b)} \right) - \int\limits_0^{\bar b} {G(b)f\left( {{w^R}(b)} \right)\frac{{\partial {w^R}(b)}}{{\partial b}}db}$.

$ \begin{array}{c} \frac{{\partial \phi }}{{\partial {\sigma_R}}} = \lambda g\left( {\bar b} \right)\frac{{\partial \bar b}}{{\partial {\sigma_R}}} - \lambda g\left( {\bar b} \right)F\left( {{w^R}(\bar b)} \right)\frac{{\partial \bar b}}{{\partial {\sigma_R}}} - \lambda G\left( {\bar b} \right)f\left( {{w^R}(\bar b)} \right)\frac{{\partial {w^R}(\bar b)}}{{\partial {\sigma_R}}} \cr \quad - \lambda G\left( {\bar b} \right)\frac{{\partial F\left( {{w^R}(\bar b)} \right)}}{{\partial {\sigma_R}}} + \lambda G\left( {\bar b} \right)f\left( {{w^R}(\bar b)} \right)\frac{{\partial {w^R}(\bar b)}}{{\partial b}}\frac{{\partial \bar b}}{{\partial {\sigma_R}}}\cr \quad + \lambda \int\limits_0^{\bar b} {G(b)\frac{{\partial f\left( {{w^R}(b)} \right)}}{{\partial {\sigma_R}}}\frac{{\partial {w^R}(b)}}{{\partial b}} db,}\end{array} $

((23))

where $\frac{{\partial F\left( {{w^R}(\bar b)} \right)}}{{\partial {\sigma_R}}} = \frac{{\partial F(w)}}{{\partial {\sigma_R}}}$ at $w = {w^R}(\bar b)$, $\frac{{\partial {w^R}(\bar b)}}{{\partial b}} = \frac{{\partial {w^R}(b)}}{{\partial b}}$ at $b = \bar b$. Clearly, $\frac{{\partial {w^R}(\bar b)}}{{\partial b}}\frac{{\partial \bar b}}{{\partial {\sigma_R}}} = \frac{{\partial {w^R}(\bar b)}}{{\partial {\sigma_R}}}$ with $\frac{{\partial {w^R}(\bar b)}}{{\partial {\sigma_R}}} = \frac{{\partial {w^R}(b)}}{{\partial {\sigma_R}}}$ at $b = \bar b$.

Hence,

$ \frac{{\partial \phi }}{{\partial {\sigma_R}}} = \lambda g\left( {\bar b} \right)\frac{{\partial \bar b}}{{\partial {\sigma_R}}}\left[ {1 - F\left( {{w^R}(\bar b)} \right)} \right] - \lambda G\left( {\bar b} \right)\frac{{\partial F\left( {{w^R}(\bar b)} \right)}}{{\partial {\sigma_R}}} + \lambda \int\limits_0^{\bar b} {G(b)\frac{{\partial f\left( {{w^R}(b)} \right)}}{{\partial {\sigma_R}}}\frac{{\partial {w^R}(b)}}{{\partial b}}} db. $

((24))

As was shown (see proof of Proposition 3), $\frac{{\partial \bar b}}{{\partial {\sigma_R}}} > 0$. Moreover, $\frac{{\partial F\left( {{w^R}(\bar b)} \right)}}{{\partial {\sigma_R}}} < 0$. Since $\int\limits_{\bar w}^\infty {f(w)dw} = 1/2$, there exists some point y, such that: $\frac{{\partial f\left( {{w^R}(b)} \right)}}{{\partial {\sigma_R}}} < 0$ for ${w^R}(b) < y$ and $\frac{{\partial f\left( {{w^R}(b)} \right)}}{{\partial {\sigma_R}}} > 0$ for ${w^R}(b) > y$.

$\frac{{\partial {w^R}(b)}}{{\partial b}} = \frac{1}{{1 + \lambda /r(1 - F({w^R}))}}$, which implies that $\frac{{\partial {w^R}(b)}}{{\partial b}}$ increases in ${w^R}(b)$ and hence in b. Hence,

$ \lambda \int\limits_0^{\bar b} {G(b)\frac{{\partial f\left( {{w^R}(b)} \right)}}{{\partial {\sigma_R}}}\frac{{\partial {w^R}(b)}}{{\partial b}}} db > \lambda G\left( {\bar b} \right)\int\limits_0^{\bar b} {\frac{{\partial f\left( {{w^R}(b)} \right)}}{{\partial {\sigma_R}}}\frac{{\partial {w^R}(b)}}{{\partial b}}} db $

((25))

The interpretation is the following: in the integral on the left side of (25), negative values have smaller weights and positive values have larger weights, hence, if we weigh all the values equally, the resulting integral (the right-hand side of (25)) would be “more negative”. Therefore, if we replace the last term in (24) with the left-hand side of (25), the resulting sum would be smaller.

Therefore, if $\lambda g\left( {\bar b} \right)\frac{{\partial \bar b}}{{\partial {\sigma_R}}}\left[ {1 - F\left( {{w^R}(\bar b)} \right)} \right] - \lambda G\left( {\bar b} \right)\frac{{\partial F\left( {{w^R}(\bar b)} \right)}}{{\partial {\sigma_R}}} + \lambda G\left( {\bar b} \right)\int\limits_0^{\bar b} {\frac{{\partial f\left( {{w^R}(b)} \right)}}{{\partial {\sigma_R}}}\frac{{\partial {w^R}(b)}}{{\partial b}}} db > 0$, then also $\lambda g\left( {\bar b} \right)\frac{{\partial \bar b}}{{\partial {\sigma_R}}}\left[ {1 - F\left( {{w^R}(\bar b)} \right)} \right] - \lambda G\left( {\bar b} \right)\frac{{\partial F\left( {{w^R}(\bar b)} \right)}}{{\partial {\sigma_R}}} + \lambda \int\limits_0^{\bar b} {G(b)\frac{{\partial f\left( {{w^R}(b)} \right)}}{{\partial {\sigma_R}}}} db > 0$. But $\lambda G\left( {\bar b} \right)\int\limits_0^{\bar b} {\frac{{\partial f\left( {{w^R}(b)} \right)}}{{\partial {\sigma_R}}}} db = \lambda G\left( {\bar b} \right)\frac{{\partial F\left( {{w^R}(\bar b)} \right)}}{{\partial {\sigma_R}}}$. Hence, $\frac{{\partial \phi }}{{\partial {\sigma_R}}} > 0$. Allowing endogenous search intensity does not change the result qualitatively: $\frac{{\partial \phi }}{{\partial {\sigma_{BR}}}} > 0$ as $\frac{{\partial {\lambda_B}({\theta_B})}}{{\partial {\sigma_{BR}}}} > 0$, where ${\sigma_{BR}}$ stands for the spread in the right tail of the wage distribution in the destination. Hence, the commuter flow unambiguously increases with the spread in the right tail of the wage distribution in the destination. However, $\frac{{\partial {\lambda_B}({\theta_B})}}{{\partial {\sigma_{AR}}}} < 0$, thus the effect of the change in the spread in the right tail in the origin is ambiguous.

In the same fashion, it can be established that the commuter flow unambiguously declines with the spread in the left tail of the wage distribution in the destination. The effect of the change in the spread in the left tail in the origin is ambiguous.

1.4 Negative Binomial Model

Let ${\lambda_{ij}} = {\mu_{ij}}{\nu_{ij}}$. If we specify ${\mu_{ij}} = {e^{x\beta }}$ and ${\nu_{ij}}$ have a Gamma distribution with $E\left[ {{\nu_{ij}}} \right] = 1$ and ${\rm var} \left[ {{\nu_{ij}}} \right] = \alpha$, then the distribution of ${y_{ij}}$ can be written as:

$ h\left( {{y_{ij}},\alpha, \mu } \right) = \frac{{\Gamma \left( {{\alpha^{ - 1}} + {y_{ij}}} \right)}}{{\Gamma \left( {{\alpha^{ - 1}}} \right)\Gamma \left( {1 + {y_{ij}}} \right)}}{\left( {\frac{{{\alpha^{ - 1}}}}{{{\alpha^{ - 1}} + {\mu_{ij}}}}} \right)^{{\alpha^{ - 1}}}}{\left( {\frac{{{\mu_{ij}}}}{{{\alpha^{ - 1}} + {\mu_{ij}}}}} \right)^{{y_{ij}}}} $

((26))

The first two moments of the negative binomial distribution are: $E\left[ {{y_{ij}}} \right] = {\mu_{ij}}$ and ${\rm var} \left[ {{y_{ij}}} \right] = {\mu_{ij}}\left( {1 + \alpha {\mu_{ij}}} \right)$. If $\alpha$ is zero then $E\left[ {{y_{ij}}} \right] = {\rm var} \left[ {{y_{ij}}} \right]$ and negative binomial is identical to the Poisson. Hence, testing $\alpha = 0$ after estimating the negative binomial is identical to testing the negative binomial specification versus the Poisson (Table 1).

1.5 Zero-Inflated Models

In zero-inflated models, zeros could be generated by two different processes. Two processes are characterized by two density functions: a binary density ${h_2}\left( \cdot \right)$ and a count density ${h_2}\left( \cdot \right)$. If the binary process generates zero (with probability ${h_1}(0)$), then ${y_{ij}} = 0$. Otherwise, ${y_{ij}}$ takes count values 0, 1, 2 … from the density ${h_2}\left( \cdot \right)$.

Hence:

$ q(y_{ij}) = \left\{\begin{array}{c} h_{1}(0) + (1 - h_{1}(0)) h_{2}(0) \quad if \quad y_{ij} = 0 \\ (1 - h_{1}(0))h_{2}(y_{ij}) \quad \quad \quad if \quad y_{ij} \geq 1. \end{array}\right. $

((27))

The likelihood function follows immediately from (27).

1.6 Data Used

The description of the IABS data set is taken from Möller and Aldashev (2006). The data on wages and wage dispersion were calculated from IABS-REG. IABS-REG is a 2% random sample from the employment register of the Federal Labour Office with regional information. The data set includes all workers, salaried employees and trainees obliged to pay social security contributions, and covers more than 80% of all employment. Public servants, minor employment and family workers are excluded (see Bender et al. 2000 for an extensive description of the data). Because of legal sanctions for misreporting, the earnings information in the data is highly reliable. Among others, IABS-REG contains variables on individual earnings and skills. The regional information is based on the employer. For the empirical analysis, the data were restricted to full-time workers of the intermediate skill group (apprenticeship completed without a university-type of education). All male and female workers selected were in employment on June 30th, 1997. For all regions the median wage and the second and eighth decile of daily earnings were calculated.

The INKAR database of the Federal Office for Building and Regional Planning contains basic geographic and demographic indicators on a regional level. The regional population used in estimation in Sect. 6 was taken from the INKAR dataset.

The data on commuting time was obtained from the Federal Statistical Office (Statistisches Bundesamt). The travel time represents average travel time by personal car between the administrative centres of the regions.